Question

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Answer 1

\[\Large {\int\limits_\alpha^\beta \left( x-\alpha \right)\left( x-\beta\right)\\ =\int\limits_\alpha^\beta x^2-\left( \alpha +\beta\right)x+\alpha \beta\\ =\left[ \frac 13x^3 -{\left( \alpha+\beta \right)\over2}x^2+\alpha \beta x\right]_\alpha^\beta\\ =\frac 16\left[ 2x^3 -3\left( \alpha+\beta \right)x^2+6\alpha \beta x\right]_\alpha^\beta}\]

Answer 2

\[\large{=\frac 16\left( 2\beta^3 -3\alpha \beta^2-3\beta^3+6\alpha \beta^2\right)-\frac 16\left( 2\alpha^3 -3\alpha^3-3\alpha^2\beta +6\alpha^2 \beta\right)\\ =\frac16\left( -\beta^3+3\alpha \beta^2 \right)-\frac16\left( -\alpha^3+3\alpha^2 \beta \right)\\=-\frac {\left( \beta-\alpha \right)^3}6}\]

Answer 3

Find area bounded by \(y=x^2\) and \(y=x+2\)

Answer 4

\[x^2-x-2\\=(x-2)(x+1)\]

\[A={(2+1)^3 \over6}=\frac92\]